What is the formula for calculating the optical output power of an LED?

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SUMMARY

The formula for calculating the optical output power of an LED is defined as Optical output power (Pout) = N × Vres, where N is a linear factor dependent on the light emission characteristics of the LED. To determine relative wall plug efficiency, one can measure the voltage drop across a resistor and the current intensity. It is essential to reference a detailed LED datasheet for accurate values, as N varies with the light emission angle and cannot be easily determined without calibration. The discussion emphasizes that for comparative efficiency measurements, the exact value of N is not necessary as it cancels out in ratio calculations.

PREREQUISITES
  • Understanding of LED characteristics and datasheets
  • Basic knowledge of electrical measurements (voltage, current)
  • Familiarity with the concept of wall plug efficiency
  • Knowledge of light emission angles and steradians
NEXT STEPS
  • Research how to read and interpret LED datasheets for optical output data
  • Learn about the concept of steradians and its impact on light measurement
  • Explore methods for calibrating light detectors for accurate power measurements
  • Investigate different types of light sources and their relative efficiencies
USEFUL FOR

Students conducting science projects on LEDs, electrical engineers, and researchers interested in optical power measurements and efficiency calculations.

daPoseidonGuy
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Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
 
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daPoseidonGuy said:
Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
Welcome to the PF.

I'd start with a good datasheet for a typical LED. There should be curves for optical output versus current. The old Hewlett Packard LED datasheets were especially complete. Can you post a link to a detailed LED datasheet that has some of the numbers you are looking for? :smile:
 
Heres a more complete post:
Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I don't know what help is. Id appreciate help a lot, thanks.
All the values I have are experimental. I am trying to calculate the relative wall plug efficiency of an LED. here's my data right now.
I do not have a data sheet and the steps I am following at this point are somewhat modeled of those shown here:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure
in the testing and data collection section.

Voltage across resistor (V) ± .01 = 2.49

Distance from photocell to light (cm) ± .05 = 4.00

Voltage across light (V) ± .01 = 5.75

Current intensity (mA) ± .01 = 360
If N is just something I have to leave as a variable, then what is the point of measuring the distance between the photocell and the light bulb? In the experiment I moved the breadboard closer or further to get 2.5v across the resistor, as that's what I understood from the procedure. Was I supposed to do that? What do I do with the value for the distance between the photocell and light bulb?
 
Last edited:
that experiment doesn't give a value for N, only compares N for LED vs incandescent lamp.
Try this datasheet..http://www.vishay.com/docs/81011/tsal6400.pdf

and see if this helps you estimate N for that particular LED

upload_2015-12-27_14-55-57.png


good opportunity to learn about steradians .
 
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From the project link:
N cannot be determined easily, as it depends on the light emission vs. angle for each source.

The project is
1. Build a light detector. Don't worry about the detector's N since we assume it is constant across all emitters.
2. Try different sources and calculate input power vs output power (efficiency) using the detector.

You don't really need to know N to determine relative efficiency of two different sources since they assume the detector's N is the same for all sources. They also assume the detector is 100% efficient, that is, it is detecting all the light from each source (or a similar fraction).
"Since you are not collecting all of the light at the light-to-voltage converter (some of the light goes off to the side), the calculation is relative."

When you take the ratio of the two sources, the N will cancel out.

In other words, you are sort of measuring light output in units of N. N goes away when you take ratios of two measurements.

If you want to measure the absolute power of a single emitter, you need to somehow calibrate the detector, which requires a known source, etc.
 
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